老师，sin=1，-1,±1时，角度分别为多少用那种kπ+1╱2π表示，cos的问题同 sin(kπ/6)=1/2sin(π/12)[cos(2k-...
k\u03c0-\u03c0\u25712\u548c k\u03c0-3\u03c0\u25712\u533a\u522bk\u503c\u786e\u5b9a\u4f1a\u76f8\u5dee\u4e00\u5708
cos(2k-1)\u03c0/12 = cos(k\u03c0/6-\u03c0/12) = cos(k\u03c0/6)cos(\u03c0/12) + sin(k\u03c0/6)sin(\u03c0/12)
cos(2k+1)\u03c0/12 = cos(k\u03c0/6+\u03c0/12) = cos(k\u03c0/6)cos(\u03c0/12) - sin(k\u03c0/6)sin(\u03c0/12)
\u4e0a\u8ff0\u4e24\u5f0f\u76f8\u51cf\uff0c\u5f97 2*sin(k\u03c0/6)sin(\u03c0/12)\uff0c
\u56e0\u6b64 1/[2sin(\u03c0/12)] * [cos(2k-1)\u03c0/12-cos(2k+1)\u03c0/12]
=1/[2sin(\u03c0/12)] * 2*sin(k\u03c0/6)sin(\u03c0/12)
= sin(k\u03c0/6)
sin x=-1，x=2kπ-π/2
sin x=±1，x=kπ+π/2
cos x=1，x=2kπ
cos x=-1，x=2kπ+π
cos x=±1，x=kπ
sinα=1时 α=2kπ+π/2
sinα=-1时 α=2kπ+ 3π/2
cosα=1时 α=2kπ
cosα=-1时 a而发=2kπ+π
绛旓細sin x=1锛x=2k蟺+蟺/2 sin x=-1锛寈=2k蟺-蟺/2 sin x=卤1锛x=k蟺+蟺/2 cos x=1锛寈=2k蟺 cos x=-1锛寈=2k蟺+蟺 cos x=卤1锛寈=k蟺
绛旓細鍥犱负sin(x)鍜宻in(y)閮芥槸姝ｅ鸡鍊硷紝鑼冨洿鍦╗-1锛1]鑼冨洿鍐咃紝鎵浠ユ牴鎹鎰 宸茬煡sin(x)sin(y)=1鍒欐湁锛歴in(x)=1,sin(y)=1鎴栬卻in(x)=-1,sin(y)=-1 鎵浠ュ彲姹傚緱 x= pi/2 + 2k*pi y=pi/2 + 2m*pi 鎴栬厁=3pi/2 + 2k*pi y=3pi/2 + 2m*pi 閭ｄ箞x+y=pi+2(k+m)pi or x...
绛旓細鍥炵瓟锛歬*180,90+k*360,270+k*360,90+180*k. cos,90+180*k,k*360,180+k*360,180*k.k鍙栨暣鏁般
绛旓細瑙ｆ瀽锛氣埖鍑芥暟f(x)=sin(wx+蠁)(w>0,0鈮は嗏墹蟺)锛寍f(0)|=1 鍗冲綋x=0鏃讹紝鍑芥暟f(x)鍙栨瀬鍊=卤1 褰揻(0)=1鏃讹紝f(x)=sin(wx+蟺/2)锛屽綋f(0)=-1鏃讹紝f(x)=sin(wx-蟺/2)鈭礷(x)鐨勫浘鍍忓叧浜庣偣M(3蟺/4,0)瀵圭О Wx+蟺/2=k蟺==>x=k蟺/w-蟺/(2w)锛涘綋k=2鏃讹紝2蟺/w-蟺...
绛旓細鈭 2sin鈭燘OA = 1 鈭 sin鈭燘OA = 1/2 鈭 鈭燘OA = 30掳鎴 150掳 褰撯垹BOA = 150掳鏃讹紝杩囩偣B浣淏E鈯杞翠簬E锛屽垯 鈭燘OE = 30掳 鈭 BE = (1/2)脳OB = 1(30掳鎵瀵圭殑鐩磋杈圭瓑浜庢枩杈圭殑涓鍗)鈭 鐐笲鐨勬í鍧愭爣涓1 鎴 -1銆傛妸鐐笲鐨勬í鍧愭爣浠ｅ叆鍦唜² + y² = 4锛屾眰寰楃偣...
绛旓細浣犲ソ锛岃В绛斿涓嬶細鍥犱负sinacosb=1 鎵浠ina鍜宑osb鍚屼负1鎴栬-1 鍚屼负1鏃讹紝a = 蟺/2 + 2k蟺 锛 b = 2k蟺锛宬鈭圸 鎵浠in锛坅 - b锛= sin蟺/2 = 1 鍚屼负-1鏃讹紝a = -蟺/2 + 2k蟺 锛 b = 蟺 + 2k蟺 鎵浠in锛坅 - b锛= sin锛-3蟺/2锛= 1 缁间笂鎵杩锛宻in锛坅 - b锛= 1 ...
绛旓細n=1鏃鍏紡鎴愮珛锛涚幇鍦ㄥ亣璁惧n-1鍏紡鎴愮珛 閭ｄ箞sinx+sin2x+sin3x+鈥︹+sinnx=sinx+sin2x+sin3x+鈥︹+sin(n-1)x+sinnx =[sin((n-1)x/2)sin(nx/2)]/sin(x/2)+sinnx =[sin((n-1)x/2)sin(nx/2)+sinnxsin(x/2)]/sin(x/2)=sin(nx/2)[sin((nx/2-x/2)+2cos(nx/2)...
绛旓細鐢遍鐭锛宻in伪*sin尾=1锛鎵浠ワ紝sin伪=sin尾=卤1 鍗砪os伪=cos尾=0 鍒檆os(伪+尾)=cos伪*cos尾-sin伪*sin尾 =0-1 =-1
绛旓細浣欏鸡鍑芥暟褰撲笖浠呭綋(x=2k蟺)鏃跺彇寰楁渶澶у1锛涘綋涓斾粎褰(x=2k蟺+蟺)鏃跺彇寰楁渶灏忓-1.鍏堣浣0锝2蟺涔嬮棿鐨勭壒娈婁笁瑙掑嚱鏁板:sin(蟺/2)=sin(90º)=1,sin(-蟺/2)=sin(-90º)=sin(270º)=-1 cos(0)=cos(0º)=1,cos(蟺)=cos(180º)=-1 鐒跺悗鍐嶅姞涓妔in(x)...
绛旓細sin(2伪)=2sin伪路cos伪 cos(2伪)=cos^2(伪)-sin^2(伪)=2cos^2(伪)-1=1-2sin^2(伪)tan(2伪)=2tan伪/[1-tan^2(伪)]路涓夊嶈鍏紡锛歴in3伪=3sin伪-4sin^3(伪)cos3伪=4cos^3(伪)-3cos伪 路鍗婅鍏紡锛歴in^2(伪/2)=(1-cos伪)/2 cos^2(伪/2)=(1+cos伪)/2 ta...
